For a continuous-time (CT) signal, the independent variable is
always enclosed by a parenthesis ( N ) (N) ( N ) x ( t ) , y ( t ) , z ( t ) , A ( x , y ) x(t), y(t), z(t), A(x, y) x ( t ) , y ( t ) , z ( t ) , A ( x , y )
For a discrete-time (DT) signal, the independent variable is always
enclosed by a brackets [ N ] [N] [ N ] x [ n ] , y [ n ] , z [ n ] , A [ m , n ] x[n], y[n], z[n], A[m, n] x [ n ] , y [ n ] , z [ n ] , A [ m , n ]
p ( t ) = v ( t ) i ( t ) = 1 R v 2 ( t ) p(t)=v(t)i(t)=\frac{1}{R}v^2(t) p ( t ) = v ( t ) i ( t ) = R 1 v 2 ( t ) ≜ is "by definition"symbol \triangleq \text{is "by definition"symbol} ≜ is "by definition"symbol
E = ∫ t 1 t 2 p ( t ) d t = ∫ t 1 t 2 1 R v 2 ( t ) d t E =\int^{t_2}_{t_1}p(t)dt=\int^{t_2}_{t_1}\frac{1}{R}v^2(t)dt E = ∫ t 1 t 2 p ( t ) d t = ∫ t 1 t 2 R 1 v 2 ( t ) d t P = E t 2 − t 1 = ∫ t 1 t 2 p ( t ) d t = ∫ t 1 t 2 1 R v 2 ( t ) d t P =\frac{E}{t_2-t_1}=\int^{t_2}_{t_1}p(t)dt=\int^{t_2}_{t_1}\frac{1}{R}v^2(t)dt P = t 2 − t 1 E = ∫ t 1 t 2 p ( t ) d t = ∫ t 1 t 2 R 1 v 2 ( t ) d t x ( t ) = { x 0 < x < 3 6 − x 3 < x < 6 E = ∫ 0 6 x ( t ) 2 d t = ∫ 0 3 ( t ) 2 d t + ∫ 3 6 ( 6 − t ) 2 d t P = E 6 − 0 = 1 6 ( ∫ 0 3 ( t ) 2 d t + ∫ 3 6 ( 6 − t ) 2 d t ) x(t)=\begin{cases}
x& 0<x<3\\
6-x& 3<x<6\\
\end{cases}\\
E=\int^{6}_{0}x(t)^2dt=\int^{3}_{0}(t)^2dt+\int^{6}_{3}(6-t)^2dt\\
P=\frac{E}{6-0}=\frac{1}{6}(\int^{3}_{0}(t)^2dt+\int^{6}_{3}(6-t)^2dt) x ( t ) = { x 6 − x 0 < x < 3 3 < x < 6 E = ∫ 0 6 x ( t ) 2 d t = ∫ 0 3 ( t ) 2 d t + ∫ 3 6 ( 6 − t ) 2 d t P = 6 − 0 E = 6 1 ( ∫ 0 3 ( t ) 2 d t + ∫ 3 6 ( 6 − t ) 2 d t ) x [ t ] = { x 0 < x < 3 6 − x 3 < x < 6 E = ∑ t = 0 3 x ( t ) 2 = ∑ t = 0 3 ( t ) 2 + ∑ t = 3 6 ( 6 − t ) 2 P = E 6 − 0 = 1 6 ( ∑ t = 0 3 ( t ) 2 + ∑ t = 3 6 ( 6 − t ) 2 ) x[t]=\begin{cases}
x& 0<x<3\\
6-x& 3<x<6\\
\end{cases}\\
E=\sum^{3}_{t=0}x(t)^2=\sum^{3}_{t=0}(t)^2+\sum^{6}_{t=3}(6-t)^2\\
P=\frac{E}{6-0}=\frac{1}{6}(\sum^{3}_{t=0}(t)^2+\sum^{6}_{t=3}(6-t)^2) x [ t ] = { x 6 − x 0 < x < 3 3 < x < 6 E = t = 0 ∑ 3 x ( t ) 2 = t = 0 ∑ 3 ( t ) 2 + t = 3 ∑ 6 ( 6 − t ) 2 P = 6 − 0 E = 6 1 ( t = 0 ∑ 3 ( t ) 2 + t = 3 ∑ 6 ( 6 − t ) 2 ) x ( t ) = E v { x ( t ) } + O d { x ( t ) } x(t)=E_v\{x(t)\}+O_d\{x(t)\}\\ x ( t ) = E v { x ( t )} + O d { x ( t )} E v { x ( t ) } = x ( t ) + x ( − n ) 2 E_v\{x(t)\}=\frac{x(t)+x(-n)}{2} E v { x ( t )} = 2 x ( t ) + x ( − n ) O d { x ( t ) } = x ( t ) − x ( − n ) 2 O_d\{x(t)\}=\frac{x(t)-x(-n)}{2} O d { x ( t )} = 2 x ( t ) − x ( − n ) ∫ − ∞ ∞ E v { x ( t ) } × O d { x ( t ) } d t = ∫ − ∞ ∞ x ( t ) + x ( − t ) 2 × x ( t ) − x ( − t ) 2 d t = ∫ − ∞ ∞ x ( t ) 2 − x ( − t ) 2 4 d t = ∫ − ∞ ∞ x ( t ) 2 4 d t − ∫ − ∞ ∞ x ( − t ) 2 4 d t = 0 \begin{aligned}\\
\int_{-\infty}^{\infty} E_v\{x(t)\}\times O_d\{x(t)\} \ dt&=\int_{-\infty}^{\infty} \frac{x(t)+x(-t)}{2}\times\frac{x(t)-x(-t)}{2} \ dt\\
&=\int_{-\infty}^{\infty}\frac{x(t)^2-x(-t)^2}{4} \ dt \\
&=\int_{-\infty}^{\infty}\frac{x(t)^2}{4} \ dt - \int_{-\infty}^{\infty}\frac{x(-t)^2}{4} \ dt\\
&=0\\
\end{aligned}\\ ∫ − ∞ ∞ E v { x ( t )} × O d { x ( t )} d t = ∫ − ∞ ∞ 2 x ( t ) + x ( − t ) × 2 x ( t ) − x ( − t ) d t = ∫ − ∞ ∞ 4 x ( t ) 2 − x ( − t ) 2 d t = ∫ − ∞ ∞ 4 x ( t ) 2 d t − ∫ − ∞ ∞ 4 x ( − t ) 2 d t = 0 unit step unit impules discrcte u [ n ] = d e f { 1 n ≥ 0 0 n < 0 u [ n − n 0 ] = d e f { 1 n 0 ≥ 0 0 n 0 < 0 u[n]\stackrel{def}{=}\begin{cases}1 & n\geq 0\\0 & n< 0\\\end{cases}\\u[n-n_0]\stackrel{def}{=}\begin{cases}1 & n_0\geq 0\\0 & n_0< 0\\\end{cases} u [ n ] = d e f { 1 0 n ≥ 0 n < 0 u [ n − n 0 ] = d e f { 1 0 n 0 ≥ 0 n 0 < 0 δ [ n ] = d e f { 1 n = 0 0 n ≠ 0 δ [ n − n 0 ] = d e f { 1 n = n 0 0 n ≠ n 0 \delta[n]\stackrel{def}{=}\begin{cases}1& n=0\\0& n\neq 0\end{cases}\\ \delta[n-n_0]\stackrel{def}{=}\begin{cases}1& n=n_0\\0& n\neq n_0\end{cases} δ [ n ] = d e f { 1 0 n = 0 n = 0 δ [ n − n 0 ] = d e f { 1 0 n = n 0 n = n 0 continouse u ( n ) = d e f { 1 n ≥ 0 0 n < 0 u(n)\stackrel{def}{=}\begin{cases}1 & n\geq 0\\0 & n< 0\\\end{cases}\\ u ( n ) = d e f { 1 0 n ≥ 0 n < 0 δ ( n ) = d e f { ∞ n = 0 0 n ≠ 0 ∫ − ∞ ∞ δ ( x ) d x = 1 \delta(n)\stackrel{def}{=}\begin{cases}\infty & n= 0\\0 & n\neq 0\\\end{cases}\\\int_{-\infty}^{\infty}\delta(x) \ dx=1\\ δ ( n ) = d e f { ∞ 0 n = 0 n = 0 ∫ − ∞ ∞ δ ( x ) d x = 1
only the current signal
y [ n ] = ( 2 x [ n ] − x [ n ] 2 ) 2 y[n]=(2x[n]-x[n]^2)^2 y [ n ] = ( 2 x [ n ] − x [ n ] 2 ) 2 only the current signal
y [ n ] = ∑ k = − ∞ n x [ k ] (accumulator) y ( t ) = ∫ − ∞ t x ( t ) d t (integral) y [ n ] = x [ n ] (delay) \begin{aligned}\\
&y[n]=\sum_{k=-\infty}^{n}x[k] & \text{(accumulator)}\\
&y(t)=\int_{-\infty}^{t} x(t) \ dt & \text{(integral)}\\
&y[n]=x[n] & \text{(delay)}\\
\end{aligned} y [ n ] = k = − ∞ ∑ n x [ k ] y ( t ) = ∫ − ∞ t x ( t ) d t y [ n ] = x [ n ] (accumulator) (integral) (delay) function is invertable
y ( t ) = 2 x ( t ) , y ( t ) − 1 = 1 2 x ( t ) y [ n ] = ∑ − ∞ n x [ n ] , y [ t ] − 1 = x [ n ] − x [ n − 1 ] \begin{aligned}\\
&y(t)=2x(t)&,&y(t)^{-1}=\frac{1}{2}x(t)\\
&y[n]=\sum_{-\infty}^{n}x[n]&,&y[t]^{-1}=x[n]-x[n-1]\\
\end{aligned} y ( t ) = 2 x ( t ) y [ n ] = − ∞ ∑ n x [ n ] , , y ( t ) − 1 = 2 1 x ( t ) y [ t ] − 1 = x [ n ] − x [ n − 1 ] only the current and past signal are relate then it is causal system
y ( t ) = x ( t − 1 ) y [ n ] = x [ n ] − x [ n − 1 ] y(t)=x(t-1)\\
y[n]=x[n]-x[n-1] y ( t ) = x ( t − 1 ) y [ n ] = x [ n ] − x [ n − 1 ] y ( t ) = x ( t + 1 ) y [ n ] = x [ n ] − x [ n + 1 ] y(t)=x(t+1)\\
y[n]=x[n]-x[n+1] y ( t ) = x ( t + 1 ) y [ n ] = x [ n ] − x [ n + 1 ] can find BIBO(bounded-input and bounded-output) in another word the function is diverage or not.
∣ x ( t ) ∣ ≤ ∞ and ∣ y ( t ) ∣ ≤ ∞ for all t |x(t)|\leq \infty \text{ and }|y(t)|\leq \infty \text{ for all t} ∣ x ( t ) ∣ ≤ ∞ and ∣ y ( t ) ∣ ≤ ∞ for all t y [ n ] = x [ n ] + x [ n + 1 ] y[n]=x[n]+x[n+1] y [ n ] = x [ n ] + x [ n + 1 ] y ( t ) = 1.01 x [ n − 1 ] y(t)=1.01x[n-1]\\ y ( t ) = 1.01 x [ n − 1 ] the function shift input will only shift and dont have any effect
, y ( t ) = t x ( t ) define y 1 ( x ) let x ( t ) = x ( t + t 0 ) , y 1 ( t ) = t x ( t + t 0 ) if y 1 ( t ) = = y ( t + t 0 ) then it is invariance , ∵ ( t + t 0 ) x ( t + t 0 ) ≠ t x ( t + t 0 ) ∴ not invariance \begin{aligned}\\
&,&&y(t)=tx(t)&\\
\text{define } y_1(x)\text{ let } x(t)=x(t+t_0)&,&&y_1(t)=tx(t+t_0)&\\
\text{if }y_1(t)==y(t+t_0) \text{then it is invariance}&,&&\because (t+t_0)x(t+t_0)\neq tx(t+t_0)& \therefore \text{not invariance}\\
\end{aligned}
define y 1 ( x ) let x ( t ) = x ( t + t 0 ) if y 1 ( t ) == y ( t + t 0 ) then it is invariance , , , y ( t ) = t x ( t ) y 1 ( t ) = t x ( t + t 0 ) ∵ ( t + t 0 ) x ( t + t 0 ) = t x ( t + t 0 ) ∴ not invariance if a y ( x ( t ) ) + b y ( x ( t ) ) = = y ( a x ( t ) + b x ( t ) ) ay(x(t))+b y(x(t))== y(ax(t)+bx(t)) a y ( x ( t )) + b y ( x ( t )) == y ( a x ( t ) + b x ( t ))
memoryless stable causal linaer time invariant y ( t ) = c o s ( x ( t ) ) y(t)=cos(x(t)) y ( t ) = cos ( x ( t )) ✅ ✅ ✅ ❌ ✅ y [ n ] = 2 x [ n ] u [ n ] y[n]=2x[n]u[n] y [ n ] = 2 x [ n ] u [ n ] ✅ ✅ ✅ ✅ ❌ y ( t ) = ∫ − ∞ t / 2 x ( u ) d u y(t)=\int_{-\infty}^{t/2} x(u) du y ( t ) = ∫ − ∞ t /2 x ( u ) d u ❌ ✅ ❌ ✅ ❌ y [ n ] = ∑ k = − ∞ n x [ k + 2 ] y[n]=\sum_{k=-\infty}^{n} x[k+2] y [ n ] = ∑ k = − ∞ n x [ k + 2 ] ❌ ✅ ❌ ✅ ✅ y ( t ) = x ( 2 − t ) y(t)=x(2-t) y ( t ) = x ( 2 − t ) ❌ ✅ ❌ ✅ ❌ y [ n ] = x [ n ] ∑ k = − ∞ ∞ δ [ n − 2 k ] y[n]=x[n]\sum_{k=-\infty}^{\infty}\delta[n-2k] y [ n ] = x [ n ] ∑ k = − ∞ ∞ δ [ n − 2 k ] ✅ ✅ ✅ ✅ ✅
j = − 1 z = r × e j θ = r × ( sin ( θ ) + j cos ( θ ) ) = r × sin ( θ ) + j r cos ( θ ) = α + j ω j=\sqrt{-1}\\
\\
\begin{aligned}
z&=r\times e^{j\theta}\\
&=r\times (\sin( \theta) +j \cos(\theta))\\
&=r\times \sin( \theta) +j r \cos(\theta)\\
&=\alpha+j\omega\\
\end{aligned}\\ j = − 1 z = r × e j θ = r × ( sin ( θ ) + j cos ( θ )) = r × sin ( θ ) + j r cos ( θ ) = α + jω x ( t ) = C e α t x(t)=Ce^{\alpha t}\\ x ( t ) = C e α t C is real C is complex a is real x ( t ) = C e α t x(t)=Ce^{\alpha t} x ( t ) = C e α t a is imaginary C = r × e j ϕ a = j w 0 x ( t ) = r × e j ( w 0 t + ϕ ) C=r\times e^{j\phi}\\a=j w_0 \\ x(t)=r\times e^{j(w_0 t+ \phi)} C = r × e j ϕ a = j w 0 x ( t ) = r × e j ( w 0 t + ϕ ) a is complex C = r 1 × e j ϕ a = α + j w 0 x ( t ) = ( r × e j ϕ ) ( e α t + j w 0 t ) x ( t ) = r ( e α t e j ( w 0 t + ϕ ) ) x ( t ) = r e α t ( cos ( w 0 t + ϕ ) + j sin ( w 0 t + ϕ ) ) C=r_1\times e^{j\phi} \\ a=\alpha +j w_0\\ \begin{aligned} x(t)&=(r\times e^{j\phi})(e^{\alpha t+j w_0t})\\ x(t)&=r(e^{\alpha t}e^{j (w_0t+\phi )})\\ x(t)&=r \ e^{\alpha t}(\cos(w_0t+\phi )+j \sin(w_0t+\phi ))\end{aligned} C = r 1 × e j ϕ a = α + j w 0 x ( t ) x ( t ) x ( t ) = ( r × e j ϕ ) ( e α t + j w 0 t ) = r ( e α t e j ( w 0 t + ϕ ) ) = r e α t ( cos ( w 0 t + ϕ ) + j sin ( w 0 t + ϕ ))
x ( t ) = r e j w 0 t fundamental period T 0 = 2 π w 0 x(t)=r \ e^{j w_0 t}\\
\text{fundamental period }T_0=\frac{2\pi}{w_0} x ( t ) = r e j w 0 t fundamental period T 0 = w 0 2 π x ( t ) = e j 2 t + e j 3 t fundamental period of e j 2 t = π fundamental period of e j 3 t = 2 π 3 LCD(Least Common Denominator) of π and 2 π 3 is 2 π fundamental period of e j 2 t + e j 3 t = 2 π x(t)=e^{j2t}+e^{j3t}\\
\text{fundamental period of }e^{j2t}=\pi\\
\text{fundamental period of }e^{j3t}=\frac{2\pi}{3}\\
\text{LCD(Least Common Denominator) of }\pi \text{ and } \frac{2\pi}{3} \text{ is } 2\pi\\
\text{fundamental period of }e^{j2t}+e^{j3t}=2\pi\\ x ( t ) = e j 2 t + e j 3 t fundamental period of e j 2 t = π fundamental period of e j 3 t = 3 2 π LCD(Least Common Denominator) of π and 3 2 π is 2 π fundamental period of e j 2 t + e j 3 t = 2 π fundamental period T 0 T_0 T 0 x [ n ] = x [ n + i % T 0 ] x[n]=x[n+i\% T_0] x [ n ] = x [ n + i % T 0 ] i i i
T 0 T_0 T 0 T 0 T_0 T 0
x [ n ] = r e j w 0 t = r e j m N n fundamental period T 0 = N fundamental frequency = 2 π N x[n]=r \ e^{j w_0 t}=r \ e^{j\frac{m}{N}n}\\
\text{fundamental period }T_0=N\\
\text{fundamental frequency }=\frac{2\pi}{N}\\ x [ n ] = r e j w 0 t = r e j N m n fundamental period T 0 = N fundamental frequency = N 2 π x [ n ] = e j ( 2 π 3 ) n + e j ( 3 π 4 ) n fundamental period of e j ( 2 π 3 ) n = 3 fundamental period of e j ( 3 π 4 ) n = 8 ∵ ( 2 π = 24 π 4 ) LCD(Least Common Denominator) of 3 and 8 is 24 fundamental period of e j ( 2 π 3 ) n + e j ( 3 π 4 ) n = 24 x[n]=e^{j(\frac{2\pi}{3})n}+e^{j(\frac{3\pi}{4})n}\\
\text{fundamental period of } e^{j(\frac{2\pi}{3})n}=3\\
\text{fundamental period of } e^{j(\frac{3\pi}{4})n}=8 \ \because(2\pi=\frac{24 \pi}{4})\\
\text{LCD(Least Common Denominator) of } 3 \text{ and } 8 \text{ is } 24\\
\text{fundamental period of }e^{j(\frac{2\pi}{3})n}+e^{j(\frac{3\pi}{4})n}=24\\ x [ n ] = e j ( 3 2 π ) n + e j ( 4 3 π ) n fundamental period of e j ( 3 2 π ) n = 3 fundamental period of e j ( 4 3 π ) n = 8 ∵ ( 2 π = 4 24 π ) LCD(Least Common Denominator) of 3 and 8 is 24 fundamental period of e j ( 3 2 π ) n + e j ( 4 3 π ) n = 24 ( f ∗ g ) ( t ) = ∫ − ∞ ∞ f ( τ ) g ( t − τ ) d τ (f*g)(t)=\int_{-\infty}^{\infty}{f(\tau)g(t-\tau) \ d\tau } ( f ∗ g ) ( t ) = ∫ − ∞ ∞ f ( τ ) g ( t − τ ) d τ ( f ∗ g ) [ n ] = ∑ k = − ∞ ∞ f [ k ] g [ n − k ] (f*g)[n]=\sum_{k=-\infty}^{\infty}{f[k]g[n-k] } ( f ∗ g ) [ n ] = k = − ∞ ∑ ∞ f [ k ] g [ n − k ] x [ n ] = { 0.5 n , 0 ≤ n 0 , e l s e h [ n ] = { 0.5 n , 0 ≤ n < 3 0 , e l s e \begin{aligned}
x[n]=&
\begin{cases}
0.5^{n} &, 0\leq n\\
0 &, else
\end{cases}\\
h[n]=&
\begin{cases}
0.5^{n} &, 0\leq n<3\\
0 &, else
\end{cases}
\end{aligned}
x [ n ] = h [ n ] = { 0. 5 n 0 , 0 ≤ n , e l se { 0. 5 n 0 , 0 ≤ n < 3 , e l se h\x x[n] 0 1 2 3 3 h[n] x [ n ] × h [ n ] x[n]\times h[n] x [ n ] × h [ n ] 1 0.5 0.25 0.125 0.0625 0 1 1 0.5 0.25 0.125 0.0625 1 0.5 0.5 0.25 0.125 0.0625 0.03125 2 0.25 0.5 0.125 0.0625 0.03125 0.015625 3 0 0 0 0 0 0 4 0 0 0 0 0 0
( x ∗ h ) [ 0 ] = x [ 0 ] × h [ 0 ] ( x ∗ h ) [ 1 ] = x [ 0 ] × h [ 1 ] + x [ 1 ] × h [ 0 ] ( x ∗ h ) [ 2 ] = x [ 0 ] × h [ 2 ] + x [ 1 ] × h [ 1 ] + x [ 2 ] × h [ 0 ] \begin{aligned}
(x*h)[0]&=x[0]\times h[0]\\
(x*h)[1]&=x[0]\times h[1]+x[1]\times h[0]\\
(x*h)[2]&=x[0]\times h[2]+x[1]\times h[1]+x[2]\times h[0]\\
\end{aligned} ( x ∗ h ) [ 0 ] ( x ∗ h ) [ 1 ] ( x ∗ h ) [ 2 ] = x [ 0 ] × h [ 0 ] = x [ 0 ] × h [ 1 ] + x [ 1 ] × h [ 0 ] = x [ 0 ] × h [ 2 ] + x [ 1 ] × h [ 1 ] + x [ 2 ] × h [ 0 ] x ( t ) ∗ h ( t ) = h ( t ) ∗ x ( t ) ∫ − ∞ ∞ x ( τ ) h ( t − τ ) d τ = ∫ − ∞ ∞ h ( τ ) x ( t − τ ) d τ \begin{aligned}
x(t)*h(t)&=h(t)*x(t)\\
\int_{-\infty}^{\infty}{x(\tau)h(t-\tau) \ d\tau }&=\int_{-\infty}^{\infty}{h(\tau)x(t-\tau) \ d\tau }\\
\end{aligned} x ( t ) ∗ h ( t ) ∫ − ∞ ∞ x ( τ ) h ( t − τ ) d τ = h ( t ) ∗ x ( t ) = ∫ − ∞ ∞ h ( τ ) x ( t − τ ) d τ x ( t ) ∗ ( h 1 ( t ) + h 2 ( t ) ) = x ( t ) ∗ h 1 ( t ) + x ( t ) ∗ h 2 ( t ) \begin{aligned}
x(t)*(h_1(t)+h_2(t))&=x(t)*h_1(t)+x(t)*h_2(t)\\
\end{aligned} x ( t ) ∗ ( h 1 ( t ) + h 2 ( t )) = x ( t ) ∗ h 1 ( t ) + x ( t ) ∗ h 2 ( t ) x ( t ) ∗ ( h 1 ( t ) ∗ h 2 ( t ) ) = ( x ( t ) ∗ h 1 ( t ) ) ∗ h 2 ( t ) \begin{aligned}
x(t)*(h_1(t)*h_2(t))&=(x(t)*h_1(t))*h_2(t)\\
\end{aligned} x ( t ) ∗ ( h 1 ( t ) ∗ h 2 ( t )) = ( x ( t ) ∗ h 1 ( t )) ∗ h 2 ( t ) y ( t ) = x ( t ) ∗ h ( t ) = ∫ − ∞ ∞ x ( t − τ ) h ( τ ) d τ = ∫ − ∞ ∞ x ( τ ) h ( t − τ ) d τ \begin{aligned}
y(t)&=x(t)*h(t)\\
&=\int_{-\infty}^{\infty} {x(t-\tau)h(\tau) } \ d \tau\\
&=\int_{-\infty}^{\infty} {x(\tau)h(t-\tau) } \ d \tau\\
\end{aligned} y ( t ) = x ( t ) ∗ h ( t ) = ∫ − ∞ ∞ x ( t − τ ) h ( τ ) d τ = ∫ − ∞ ∞ x ( τ ) h ( t − τ ) d τ y ( t ) = x ( t ) ∗ h ( t ) y ( t − k ) = x ( t − k ) ∗ h ( t ) \begin{aligned}
y(t)&=x(t)*h(t)\\
y(t-k)&=x(t-k)*h(t)\\
\end{aligned} y ( t ) y ( t − k ) = x ( t ) ∗ h ( t ) = x ( t − k ) ∗ h ( t ) if ∣ x ( t ) ∣ is bounded, then ∣ y ( t ) ∣ is also bounded. Sufficient condition for a continuous-time LTI system to be stable \text{if }|x(t)|\text{ is bounded, then }|y(t)|\text{ is also bounded.}\\
\text{Sufficient condition for a continuous-time LTI system to be stable} if ∣ x ( t ) ∣ is bounded, then ∣ y ( t ) ∣ is also bounded. Sufficient condition for a continuous-time LTI system to be stable y [ n ] = h [ n ] ∗ u [ t ] = ∫ − ∞ ∞ u [ n − τ ] h [ τ ] d τ = ∫ − ∞ n h [ τ ] d τ h [ n ] = y [ n ] − y [ n − 1 ] \begin{aligned}
y[n]&=h[n]*u[t]\\
&=\int_{-\infty}^{\infty}u[n-\tau]h[\tau] \ d \tau \\
&=\int_{-\infty}^{n} h[\tau] \ d \tau \\
\end{aligned}\\
h[n]=y[n]-y[n-1] y [ n ] = h [ n ] ∗ u [ t ] = ∫ − ∞ ∞ u [ n − τ ] h [ τ ] d τ = ∫ − ∞ n h [ τ ] d τ h [ n ] = y [ n ] − y [ n − 1 ] y ( t ) = h ( t ) ∗ u ( t ) = ∑ − ∞ ∞ u ( t − τ ) h ( τ ) d τ = ∑ − ∞ t h ( τ ) d τ h ( t ) = y ′ ( t ) \begin{aligned}
y(t)&=h(t)*u(t)\\
&=\sum_{-\infty}^{\infty}u(t-\tau)h(\tau) \ d \tau \\
&=\sum_{-\infty}^{t}h(\tau) \ d \tau \\
\end{aligned}\\
h(t)=y'(t) y ( t ) = h ( t ) ∗ u ( t ) = − ∞ ∑ ∞ u ( t − τ ) h ( τ ) d τ = − ∞ ∑ t h ( τ ) d τ h ( t ) = y ′ ( t ) The response of an LTI system to a eigen function is the same eigen function with only a change in amplitude(eigen value)
x ( t ) → H ( t ) x ( t ) x ( t ) is eigenfunction H ( t ) is eigen value x(t) \rightarrow H(t)x(t)\\
x(t)\text{ is eigenfunction }\\
H(t)\text{ is eigen value}\\ x ( t ) → H ( t ) x ( t ) x ( t ) is eigenfunction H ( t ) is eigen value x ( t ) = e s t y ( t ) = h ( t ) ∗ x ( t ) = ∫ − ∞ ∞ h ( τ ) x ( t − τ ) d t = ∫ − ∞ ∞ h ( τ ) e s ( t − τ ) d t = e s t ∫ − ∞ ∞ h ( τ ) e − s τ d t H ( t ) = ∫ − ∞ ∞ h ( τ ) e − s τ d t x(t)=e^{st}\\
\begin{aligned}
y(t)&=h(t)*x(t)\\
&=\int_{-\infty}^{\infty}h(\tau)x(t-\tau) \ dt\\
&=\int_{-\infty}^{\infty}h(\tau)e^{s(t-\tau)} \ dt\\
&=e^{st}\int_{-\infty}^{\infty}h(\tau)e^{-s\tau} \ dt\\
\end{aligned}\\
H(t)=\int_{-\infty}^{\infty}h(\tau)e^{-s\tau} \ dt x ( t ) = e s t y ( t ) = h ( t ) ∗ x ( t ) = ∫ − ∞ ∞ h ( τ ) x ( t − τ ) d t = ∫ − ∞ ∞ h ( τ ) e s ( t − τ ) d t = e s t ∫ − ∞ ∞ h ( τ ) e − s τ d t H ( t ) = ∫ − ∞ ∞ h ( τ ) e − s τ d t x ( t ) = e j 2 t y ( t ) = e j 2 ( t − 3 ) = e j 6 e j 2 t H ( t ) = ∫ − ∞ ∞ δ ( τ − 3 ) e − s τ d t = e − 3 s x(t)=e^{j2t}\\
y(t)=e^{j2(t-3)}=e^{j6}e^{j2t}\\
H(t)=\int_{-\infty}^{\infty}\delta(\tau-3)e^{-s\tau} \ dt=e^{-3s}\\ x ( t ) = e j 2 t y ( t ) = e j 2 ( t − 3 ) = e j 6 e j 2 t H ( t ) = ∫ − ∞ ∞ δ ( τ − 3 ) e − s τ d t = e − 3 s x ( t ) ∗ δ ( t − t 0 ) = x ( t − t 0 ) x [ t ] ∗ δ [ t − t 0 ] = x [ t − t 0 ] x(t) *\delta(t-t_0) = x(t-t_0)\\
x[t] *\delta[t-t_0] = x[t-t_0]\\ x ( t ) ∗ δ ( t − t 0 ) = x ( t − t 0 ) x [ t ] ∗ δ [ t − t 0 ] = x [ t − t 0 ] x [ n ] ∗ δ [ t − t 0 ] = x [ t − t 0 ] x[n] * \delta[t-t_0] = x[t-t_0] x [ n ] ∗ δ [ t − t 0 ] = x [ t − t 0 ] y [ n ] = ∑ m − ∞ n x [ m ] y[n] = \sum_{m-\infty}^{n} x[m] y [ n ] = m − ∞ ∑ n x [ m ] ∫ T 0 e j w 0 t d t \int_{T_0}e^{jw_0t} \ dt ∫ T 0 e j w 0 t d t ∫ T 0 e j w 0 t d t \int_{T_0}e^{jw_0t} \ dt ∫ T 0 e j w 0 t d t